The differential equation $yy''=(y')^2$ is given. Then by the following identity from the chain rule, $$p=y'=\frac{dy}{dx},\quad y''=\frac{dp}{dx}=\frac{dp}{dy}\frac{dy}{dx}=p\frac{dp}{dy},$$ the equation becomes $yp\frac{dp}{dy}=p^2$.
I don't understand how the second identity is derived, $y''=\frac{dp}{dx}=\frac{dp}{dy}\frac{dy}{dx}=p\frac{dp}{dy}$ by the chain rule. How can the function $y'=p$ be a function of both $x,y$? (from $\frac{dp}{dx},\frac{dp}{dy}$) I would like detailed explanation for the second identity.
The idea is to replace the independent variable from $x$ to $y$. This only works on segments where the solution is monotonous, so constant solutions have to be discussed separately.
Then with $y'(x)=p(y(x))$ the chain rule indeed gives $$y''(x)=p'(y(x))y'(x)=p'(y(x))p(y(x)).$$ After replacing $y'(x),y''(x)$ all components only depend on $x$ indirectly through $y(x)$, so the variable replacement is possible and results in the claimed equation.
This is mostly a visual effect, allowing to better see the structure of the equation. The next step of separation works equally well on the original equation $$ \frac{y''(x)}{y'(x)}=\frac{y'(x)}{y(x)}\implies y'(x)=Cy(x), $$ or $$ \frac{d}{dx}\frac{y'(x)}{y(x)}=\frac{y(x)y''(x)-y'(x)^2}{y(x)^2}=0. $$